Tolerance stack help needed.
Tolerance stack help needed.
(OP)
I need some help here to confirm if I am on the right track or not. I would like to find out if the X min/ X max dimensions (see sketch attached) are correct calculated.
I realize none of the fine gentlemen on this form do not work for me, therefore is more like a favor to ask for someone who has strong knowledge in those kind of calculations (or even maybe some kind of reliable tolerance stackup software Vis-VSA, 3DCS , CeTol or other).
I am trying to understand the concept (how the calculations should be done CORRECTLY) otherwise I will be only a “number generator” and my calculations are worthless sheet of paper.
Note: do not consider the form error on the datum feature A. I know does affect the calculation, but I am not there yet. (I have to walk before I can run)
I got
X min.: .057
and
X max.: .076
(should be no surprise if I am telling you that not everybody around here agrees with those numbers).
Thank you
I realize none of the fine gentlemen on this form do not work for me, therefore is more like a favor to ask for someone who has strong knowledge in those kind of calculations (or even maybe some kind of reliable tolerance stackup software Vis-VSA, 3DCS , CeTol or other).
I am trying to understand the concept (how the calculations should be done CORRECTLY) otherwise I will be only a “number generator” and my calculations are worthless sheet of paper.
Note: do not consider the form error on the datum feature A. I know does affect the calculation, but I am not there yet. (I have to walk before I can run)
I got
X min.: .057
and
X max.: .076
(should be no surprise if I am telling you that not everybody around here agrees with those numbers).
Thank you





RE: Tolerance stack help needed.
And I don't think that form error of datum feature will make any difference in results.
RE: Tolerance stack help needed.
To greenimi: did anyone "around there" suggested those numbers?
RE: Tolerance stack help needed.
As a matter of fact: YES. (two of .068’s):
X max. [(.505+.002) – (.372-.001)]/2 = .068
Other numbers for X max: .071 and .064. Go figure………
(X min. is more in agreement: .057)
I used virtual boundaries and resultant condition boundaries:
X max. .517-.365 = .152; .152/2 = .076
(.517 = .505 +.012 pos at LMC)
(.365 = .372-.007 pos at LMC)
I feel somewhat safe (99.999999%, confidence level) now that pmarc got my numbers:):) !!!!
RE: Tolerance stack help needed.
Question to CH:
Did you take datum feature shift into account?
RE: Tolerance stack help needed.
Did you take simultaneous requirement into account?
RE: Tolerance stack help needed.
RE: Tolerance stack help needed.
Yes, I did simultaneous requirement into account.
RE: Tolerance stack help needed.
There was a tip on Tec-Ease named "Datum shift is not bonus". Unfortunately free tips are not free anymore.
RE: Tolerance stack help needed.
Besides, according to the drawing datum feature simulator pin is Ø.248, right? So, simply picture the datum feature A having its actual mating envelope = exactly Ø.248. And here you have the contact between the datum feature and its simulator.
RE: Tolerance stack help needed.
Note: No datum shift because of the default SIM. REQ.!
RE: Tolerance stack help needed.
I get this from
-0.186 (radius of inner diameter)
+0.0005 (half of inner diameter's stated position tol)
+0.003 (half of inner diameter's max bonus tol)
+0.001 (half of outer diameter's stated position tol)
+0.0025 (half of outer diameter's max bonus tol)
+0.2525 (radius of outer diameter)
I know this may be a different approach than what some of you used. And no, I didn't use any datum shift. But since some of you get the max being .076, and some get .068, I thought I'd throw my vote in; let me know if I've goofed somewhere :)
John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
RE: Tolerance stack help needed.
When you use dia .378 (which is MMC), you can't have bonus! Same applies for dia .495 (which is MMC). Btw, you should use dia .495 for min. calculation.
RE: Tolerance stack help needed.
The pin (.372 to .378) has an MMC virtual diameter of .379; Since the smallest pin diameter is confined to the virtual diameter, if the pin is the smallest size (.372) one side can be offset by .379/2 = .1895; the amount remaining on the other side is .1825.
The MMC virtual condition of the hole is (.495-.002) which has a radius of .2465. This is the closest the hole can approach the axis. The largest hole is .505. The farthest radius will be .505 -.2465 = .2585
The difference is .076, which is the largest gap.
The smallest gap is 0.057 is the difference in MMC virtual condition radii.
RE: Tolerance stack help needed.
Dave -- I think I follow that, but let me sleep on it tonight. I'm just trying to reconcile my "Krulikowski" method with your VC method.
John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
RE: Tolerance stack help needed.
How about this?
-0.186 (radius of inner diameter)
+0.0005 (half of inner diameter's stated position tol)
+0.003 (half of inner diameter's max bonus tol)
+0.001 (half of outer diameter's stated position tol)
+0.005 (half of outer diameter's max bonus tol)
+0.2525 (radius of outer diameter)
You simply used wrong number for the half of outer diameter's max bonus tol (5th line in the stack).
RE: Tolerance stack help needed.
That's what happens when I do simple math late at night :)
John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
RE: Tolerance stack help needed.
If i take form error into account both for smaller diameter and larger diameter
( obviously for maximum value of x) then what my results will be?
RE: Tolerance stack help needed.
If i take form error into account both for smaller diameter and larger diameter
( obviously for maximum value of x) then what my results will be?
RE: Tolerance stack help needed.
As pmarc stated (and personally I believe him -- he has a loooooooooooog record of being right)
“And I don't think that form error of datum feature will make any difference in results. “24 Mar 14 13:19
Pmarc,
Please don’t let me down.....
RE: Tolerance stack help needed.
I don't think waqasmalik meant form error of datum feature. I believe he meant form errors of both toleranced features, that is Ø.372-.378 and Ø.495-.505.
waqasmalik,
The results will not change when you take these form errors into account. Because the simultaneous requirement applies between both features (hence no datum feature shift is available), the only thing that needs to be done in order to find Xmin and Xmax is calculating:
1. Radial distance between virtual conditions of both features (Xmin):
Xmin = [(.495-.002)-(.378+.001)]/2 = .057
These two virtual conditions represent extreme boundaries that can never be violated by material of the features.
2. Radial distance between resultant conditions of both features (Xmax):
Xmax = [(.505+.002+.010)-(.372-.001-.006)]/2 = .076
These two resultant conditions represent extreme boundaries within which the material of the features must always be.
RE: Tolerance stack help needed.
As usual, I agree with pmarc.
When calculating tolerance stackups, maximum values for things like misalignment often occur when the features are at LMC, have perfect form, and are perfectly oriented. Introducing form error adds material and actually decreases the possible misalignment. In other words, the imperfect form (or imperfect orientation) doesn't make the situation more extreme.
In this example, the maximum value of X occurs when the tolerance ID feature and the toleranced OD feature both have maximum misalignment, but in opposite directions.
Evan Janeshewski
Axymetrix Quality Engineering Inc.
www.axymetrix.ca
RE: Tolerance stack help needed.
Let say, the groove has less than perfect shape (within the tolerances of course).
How do you measure X? from B to C? from A to D? from A perpendicular to (CD)? from C perpendicular to (AB)? from A to D but parallel to upper face? lower face?
Just a thought.